Simplify the exponential equation
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I have an equation, which is in exponential format (y = A.exp(B.x)) where A and B are polynomial of order 3 which is a function of z, I have given the equation below. Is it possible to simplify this equation, by simplify means can we shorten this equation or can we put this equation in much more presentable way.
Simplify command is anyway not working with this.
y = exp(x*(0.1852*z^3 - 0.7827*z^2 + 0.8524*z + 0.8605))*(- 0.1061*z^3 + 0.23*z^2 + 0.5244*z + 0.0225)
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Rik
le 4 Mai 2023
How would you manually simplify this? I don't personally think changing the shape of the function makes it much simpler.
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Walter Roberson
le 4 Mai 2023
No. That is the most compact version that you can reasonably expect.
You can break it into pieces, but then what?
y = str2sym('exp(x*(0.1852*z^3 - 0.7827*z^2 + 0.8524*z + 0.8605))*(- 0.1061*z^3 + 0.23*z^2 + 0.5244*z + 0.0225)')
ch = children(y)
syms A B
Aval = ch{2};
Bval = children(children(ch{1},1),2);
nice_y = [sym('y') == subs(y, {Aval, Bval}, {A, B});
A == Aval;
B == Bval];
nice_y
But you probably could have done that by not creating y in that form in the first place, and using your existing A and B.
3 commentaires
Walter Roberson
le 4 Mai 2023
The values you get out of my code are polynomial functions.
If the task is to extract the polynomials from an expression of that particular form, then the task is do-able using code similar to the above. In practice I would probably make it more robust, perhaps using functions such as findSymType or related to be sure that I had picked the appropriate parts, instead of relying on positions. (The exact decomposition can get a bit weird depending on the sign of the z^1 coefficient in A)
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